### Video Transcript

Solve log to the base two of log to
the base three of π₯ squared minus eight π₯ equals one where π₯ is in the set of
real numbers.

So the first thing we can do with
this problem is make sure that weβve got a log to the same base on each side of the
equation. And we can do that by applying one
of our log rules. And that is that log to the base π
of π equals one. So, therefore, we can say that log
to the base two of log to the base three of π₯ squared minus eight π₯ equals log to
base two of two because, as we said, one is just the same as log to the base two of
two.

So now the reason weβve done this
is because what weβve got is log to the same base on the left- and right-hand
sides. So, what we can do is equate our
arguments. So, we can say that log to the base
three of π₯ squared minus eight π₯ is equal to two. So now, to solve for π₯, what we
can do first is actually rearrange from logarithmic to exponent form. So if we have log to the base π of
π equals π₯, then we say that π to the power of π₯ is gonna be equal to π.

So therefore, if we identify our
π, π, and π₯, well then we can rewrite our equation as three squared equals π₯
squared minus eight π₯, which is gonna give us nine equals π₯ squared minus eight
π₯. So then, what weβre gonna do is
subtract nine from each side of the equation. So, we get zero equals π₯ squared
minus eight π₯ minus nine. So now what we need to do is solve
this quadratic.

Well, we can solve the quadratic by
factoring. And if we do, we get zero equals π₯
minus nine multiplied by π₯ plus one. So therefore, we can say that π₯ is
gonna be equal to nine or negative one. So great, and these both are going
to be in our solution set. Well, you might think, βWell, no,
it canβt be because we know that the argument must be positive and not equal to
one.β So therefore, we cannot have the
negative value. However, that is not the case in
this particular problem because if we take a look at the argument, the argument is
not π₯. The argument is π₯ squared minus
eight π₯, well, the argument of one of our logarithms.

And what weβre gonna do to show
that is actually substitute in π₯ equals negative one into this argument. Well, if we do that, we get
negative one all squared minus eight multiplied by negative one, which is gonna give
us one plus eight because negative one squared is just one. And if you subtract a negative,
itβs the same as adding a positive. Well, this gives us the value of
nine, which is both positive and not equal to one. So therefore, it satisfies the
conditions that we have for our argument. So therefore, we can say that the
solution set for our equation is nine and negative one.